The eigenspace of \(T, \lambda\) is the set of all eigenvectors of \(T\) corresponding to \(\lambda\), plus the \(0\) vector.

## constituents

- \(T \in \mathcal{L}(V)\)
- \(\lambda \in \mathbb{F}\), an eigenvalue of \(T\)

## requirements

\begin{equation} E(\lambda, T) = \text{null}\ (T - \lambda I) \end{equation}

i.e. all vectors such that \((T- \lambda I) v = 0\).

where, \(E\) is an eigenspace of \(T\).

## additional information

### sum of eigenspaces is a direct sum

\(E(\lambda_{1}, T) + … + E(\lambda_{m}, T)\) is a direct sum.

### dimension of sum of eigenspaces is smaller than or equal to the dimension of the whole space

A correlate of the above is that:

\begin{equation} \dim E(\lambda_{1}, T) + … + \dim E(\lambda_{m}, T) \leq \dim V \end{equation}

Proof:

Recall that:

\begin{equation} \dim E(\lambda_{1}, T) + … + \dim E(\lambda_{m}, T) = \dim (E(\lambda_{1}, T) \oplus … \oplus E(\lambda_{m}, T) ) \end{equation}

Now, the sum of subspaces is the smallest subspace, so \(\dim (E(\lambda_{1}, T) \oplus … \oplus E(\lambda_{m}, T) ) \leq \dim V\).

And hence:

\begin{equation} \dim E(\lambda_{1}, T) + … + \dim E(\lambda_{m}, T) \leq \dim V \end{equation}

as desired. \(\blacksquare\)